I want to compute the integral $$2\pi\int f(x) \sqrt{1+f'(x)^2} dx$$ where $f(x)=\dfrac{1}{e^x}$.

I used maple and I found that the answer is: $$\pi e^{-2x} \left[e^{2x} \arctan\left(\sqrt{e^{2x}-1}\right) - \sqrt{e^{2x}-1}\right] $$ but I can't find a way to prove it on the paper. Any help would be apreciated.

well, if you substitute with $u=e^{-x}$ you also get $\frac{du}{dx}=-e^{-x} \Rightarrow -du = e^{-x} dx$ so you get $$ -2 \pi \int \sqrt{1+ u^2} du $$ if we set $u = tan(t)$ then we'll get $$ -2 \pi \int \frac{\sqrt{1+ tan(t)^2}}{cos(t)^2} dt $$ But how is this easier than the previous one?
–
BelialDec 4 '12 at 15:21